Tags Electromagnetic Induction

Electromagnetic Induction

Get insights from 123 questions on Electromagnetic Induction, answered by students, alumni, and experts. You may also ask and answer any question you like about Electromagnetic Induction

Follow Ask Question

123

Questions

Discussions

Active Users

Followers

New answer posted

6 months ago

Electromagnetic Induction physics ncert exemplar solutions class 12th chapter two

The dimension of mutual inductance is:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$e = - M \frac{d i}{d t}$

$M = \frac{e}{d i / d t} = M L^{2} T^{- 2} A^{- 2}$

0 0

New answer posted

6 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

In a coil of resistance $8 Ω,$ the magnetic flux due to an external magnetic field varies with time as $ϕ = \frac{2}{3} (9 - t^{2}) .$ The value of total heat produced in the coil, till the flux becomes zero, will be_______J.

0 Follower 10 Views

alok kumar singh

Contributor-Level 10

$? = \frac{2}{3} (9 - t^{2}) = 0$

t = 3 sec

$E m f = \frac{- d ?}{d t} = \frac{4 t}{3}$

$l = \frac{E m f}{R} = \frac{\frac{4}{3} t}{8} = \frac{t}{6}$

Heat = $\int l^{2} R d t = \int_{0}^{3} \frac{t^{2}}{36} * 8 d t = 2 J$

0 0

New answer posted

6 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

An aeroplane, with its wings spread 10m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 * 10^-4Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

B_v = B sin 60°

$B_{v} = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2}$

$E m f = B_{v} * v * l = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2} * 180 * \frac{5}{18} * 1 = 108.25 * 1 0^{- 3} v o l t s$

0 0

New answer posted

7 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

A transformer operating at primary voltage 8 kV and secondary voltage 160V serves a load of 80 k.W. Assuming the transformer to be ideal with purly resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

0 Follower 10 Views

Vishal Baghel

Contributor-Level 10

$V_{P} = 8 k V, V_{S} = 160 v P = 80 k W$

(Purely resistive load)

$P = V I = \frac{V_{P}^{2}}{R_{P}} = \frac{V_{S}^{2}}{R_{S}}$

$R_{S} = \frac{V_{S}^{2}}{P} = \frac{160 * 160}{80 * 1 0^{3}} = 320 * 1 0^{- 3} = 0.32 Ω$

0 0

New answer posted

7 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

A coil of inductance 1H and resistance 100 $Ω$ is connected to a battery of 6.V. Determine approximately:

(A)The time elapsed before the current acquires half of its steady – state value.

(B)The energy stored in the magnetic field associated with the coil at instant 15ms after the circuit is switched on. (Given In 2 = 0.693, e3/2 = 0.25)

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

L = 1H, R = 100 $Ω$

$A s, i = i_{0} e^{- t / τ}$

$F o r i = \frac{i_{0}}{2} \Rightarrow \frac{i_{0}}{2} = i_{0} e^{- t / τ}$

$\Rightarrow - l n 2 = - t / τ$

$i = \frac{6}{100} e^{- \frac{15}{1 / 100}} = . 06 e^{- 1500 * 1 0^{- 3}} = 0.06 e^{- 1.5} = 0.06 * 0.25 = . 0.015 - A$

So, $U = \frac{1}{2} L i^{2} = \frac{1}{2} * 1 * {(15 * 1 0^{- 3})}^{2} = \frac{1}{2} (225) * 1 0^{- 6} = 112.5 * 1 0^{- 6} J = 0.1125 m J$

0 0

New answer posted

7 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 * 10^-³ T and the angle of dip is 45°. The emf induced in the rod is………….mV.

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

$t a n δ = \frac{B_{v}}{B_{h}}$

$B v = B_{h} t a n 45$

= 4 * 10^-³

emf induced in the rod is = $B_{v} v l$

$= 4 * 1 0^{- 3} * 20 * \frac{20}{100}$

$\begin{array}{l} = 16 * 1 0^{- 3} \\ = 16 m v \end{array}$

0 0

New question posted

8 months ago

Electromagnetic Induction Class 12th

7.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

0 Follower 3 Views

New answer posted

8 months ago

Electromagnetic Induction Class 12th

6.17 A line charge l per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B = – B₀ k (r $\leq$ a; a < R)

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

6.17 Line charge per unit length = $λ$ = $\frac{T o t a l c h a r g e}{L e n g t h}$ = $\frac{Q}{2 π r}$

Where r = distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, $\overset{?}{B}$ = $- B_{0} \overset{?}{k}$

At a distance r, the magnetic force is balanced by the centripetal force. i.e.

BQ $v$ = $\frac{M v^{2}}{r}$ , where v = linear velocity of the wheel. Then,

B $* 2 λ π r$ = $\frac{M v}{r}$

v = $\frac{2 B λ π r^{2}}{M}$

Angular velocity, $ω = \frac{v}{R}$ = $\frac{2 B λ π r^{2}}{M R}$

For r $\leq a \leq R, w e g e t$ $ω = - \frac{2 B_{0} λ π a^{2}}{M R} \overset{?}{k}$

0 0

New answer posted

8 months ago

Electromagnetic Induction Class 12th

6.16 (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

0 Follower 25 Views

Payal Gupta

Contributor-Level 10

6.16 Let us take a small element dy in the loop, at a distance y from the long straight wire.

Magnetic flux associated with element dy, d $\emptyset$ = BdA, where

dA = Area of the element dy = a dy

Magnetic flux at distance y, B = $\frac{μ_{0} I}{2 π y}$ , where

I = current in the wire

$μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

Therefore,

d $\emptyset$ = $\frac{μ_{0} I}{2 π y}$ a dy = $\frac{μ_{0} I a}{2 π}$ $\frac{d y}{y}$

$\emptyset = \frac{μ_{0} I a}{2 π} \int \frac{d y}{y}$

Now from the figure, the range of y is x to x+a. Hence,

$\emptyset = \frac{μ_{0} I a}{2 π} \int_{x}^{x + a} \frac{d y}{y}$ = $\frac{μ_{0} I a}{2 π} {[\log_{e} ? y]}_{x}^{a + x}$ = $\frac{μ_{0} I a}{2 π} \log_{e} ? (\frac{a + x}{x})$

For mutual inductance M, the flux is given as

$\emptyset = M I$ . Hence

$M I = \frac{μ_{0} I a}{2 π} \log_{e} ? (\frac{a + x}{x})$

M = $\frac{μ_{0} a}{2 π} \log_{e} ? (\frac{a}{x} + 1)$

Emf induced in the loop, e = Bav

= ( $\frac{μ_{0} I}{2 π x}$ ) $*$ av

For I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s

...more

6.16 Let us take a small element dy in the loop, at a distance y from the long straight wire.

Magnetic flux associated with element dy, d $\emptyset$ = BdA, where

dA = Area of the element dy = a dy

Magnetic flux at distance y, B = $\frac{μ_{0} I}{2 π y}$ , where

I = current in the wire

$μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

Therefore,

d $\emptyset$ = $\frac{μ_{0} I}{2 π y}$ a dy = $\frac{μ_{0} I a}{2 π}$ $\frac{d y}{y}$

$\emptyset = \frac{μ_{0} I a}{2 π} \int \frac{d y}{y}$

Now from the figure, the range of y is x to x+a. Hence,

$\emptyset = \frac{μ_{0} I a}{2 π} \int_{x}^{x + a} \frac{d y}{y}$ = $\frac{μ_{0} I a}{2 π} {[\log_{e} ? y]}_{x}^{a + x}$ = $\frac{μ_{0} I a}{2 π} \log_{e} ? (\frac{a + x}{x})$

For mutual inductance M, the flux is given as

$\emptyset = M I$ . Hence

$M I = \frac{μ_{0} I a}{2 π} \log_{e} ? (\frac{a + x}{x})$

M = $\frac{μ_{0} a}{2 π} \log_{e} ? (\frac{a}{x} + 1)$

Emf induced in the loop, e = Bav

= ( $\frac{μ_{0} I}{2 π x}$ ) $*$ av

For I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s

e = $\frac{4 π * 10^{- 7} * 50 * 0.1 * 10}{2 π * 0.2}$ = 5 $* 10^{- 5}$ V

less

0 0

New answer posted

8 months ago

Electromagnetic Induction Class 12th

6.15 An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^–3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

6.15 Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 ${c m}^{2}$ = 25 $* 10^{- 4}$ $m^{2}$

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flowing time, t = $10^{- 3}$ s

We know average back emf, e = $\frac{d \emptyset}{d t}$ …………………. (1)

Where $d \emptyset =$ Change in flux = NAB ……………… .(2)

B = Magnetic field strength = $μ_{0} \frac{N I}{l}$ ………………. (3)

Where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ T m $A^{- 1}$

From equation (1) and (2), we get

e = $\frac{N A B}{d t}$ = $\frac{N A}{d t} * μ_{0} \frac{N I}{l}$ = $\frac{μ_{0} N^{2} A I}{l t}$ = $\frac{4 π * 10^{- 7} * 500^{2} * 25 * 10^{- 4} * 2.5}{0.3 * 10^{- 3}}$ = 6.544 V

Hence the average back emf induced in the solenoid is 6.5 V

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
686k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Electromagnetic Induction

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience